package Tree;

/**
 * @BelongsProject: SeniorArchitect-LeetCode
 * @BelongsPackage: Tree
 * @Author: zhuangxiaoyan
 * @CreateTime: 2023-10-24  22:31
 * @Description: TODO
 * @Version: 1.0
 */
public class 二叉树的最小深度111 {
    public int minDepth2(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int left = minDepth2(root.left);
        int right = minDepth2(root.right);
        if (left == 0) {
            return right + 1;
        } else if (right == 0) {
            return left + 1;
        }
        return Math.min(left, right) + 1;
    }

    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        //这道题递归条件里分为三种情况
        //1.左孩子和有孩子都为空的情况，说明到达了叶子节点，直接返回1即可
        if (root.left == null && root.right == null) {
            return 1;
        }
        //2.如果左孩子和由孩子其中一个为空，那么需要返回比较大的那个孩子的深度
        int m1 = minDepth(root.left);
        int m2 = minDepth(root.right);
        //这里其中一个节点为空，说明m1和m2有一个必然为0，所以可以返回m1 + m2 + 1;
        if (root.left == null) {
            return m2 + 1;
        }
        if (root.right == null) {
            return m1 + 1;
        }
        //3.最后一种情况，也就是左右孩子都不为空，返回最小深度+1即可
        return Math.min(m1, m2) + 1;
    }

    public int minDepth3(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftDepth = minDepth3(root.left);
        int rightDepth = minDepth3(root.right);
        //1.如果左孩子和右孩子有为空的情况，直接返回m1+m2+1
        //2.如果都不为空，返回较小深度+1
        return root.left == null || root.right == null ? leftDepth + rightDepth + 1 : Math.min(leftDepth, rightDepth) + 1;
    }

    class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        public TreeNode(int val) {
            this.val = val;
        }

        public TreeNode(int val, TreeNode left, TreeNode right) {
            this.val = val;
            this.left = left;
            this.right = right;
        }
    }
}
